Chọn phương án B.

$$\sin x=\dfrac{\sqrt{3}}{2}\Leftrightarrow\left[\begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\ x=\dfrac{2\pi}{3}+k2\pi\end{array}\right.\,(k\in\mathbb{Z})$$
\(\blacksquare\) Với \(x=\dfrac{\pi}{3}+k2\pi\): $$\begin{eqnarray*}
0\leq&x&\leq2\pi\\
\Leftrightarrow0\leq&\dfrac{\pi}{3}+k2\pi&\leq2\pi\\
\Leftrightarrow-\dfrac{\pi}{3}\leq&k2\pi&\leq\dfrac{5\pi}{3}\\
\Leftrightarrow-\dfrac{1}{6}\leq&k&\leq\dfrac{5}{6}
\end{eqnarray*}$$Vì \(k\in\mathbb{Z}\) nên \(k=0\).

\(\blacksquare\) Với \(x=\dfrac{2\pi}{3}+k2\pi\): $$\begin{eqnarray*}
0\leq&x&\leq2\pi\\
\Leftrightarrow0\leq&\dfrac{2\pi}{3}+k2\pi&\leq2\pi\\
\Leftrightarrow-\dfrac{2\pi}{3}\leq&k2\pi&\leq\dfrac{4\pi}{3}\\
\Leftrightarrow-\dfrac{1}{3}\leq&k&\leq\dfrac{2}{3}
\end{eqnarray*}$$Vì \(k\in\mathbb{Z}\) nên \(k=0\).

Vậy phương trình đã cho có \(2\) nghiệm trên đoạn \(\left[0;2\pi\right]\).